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Please help - An ideal gas problem involving work on a mass

  1. Apr 17, 2014 #1
    Please help -- An ideal gas problem involving work on a mass

    1. An ideal gas has a heat capacity of 20 J/(mol*K). One mole of the gas is in a cylinder and absorbs 1000 J of heat and lifts a 10 kg mass a vertical distance of 2 m while expanding. If the initial temperature of the gas is 300 K, what is the approximate final temperature of the gas?



    2. Q = mcΔT, ΔE = Q-W



    3. I thought this would be straightforward, but I'm not getting the correct answer of 340K. I thought that I could set Q=1000 J, plug in m and c, and solve for Tf. In this case, Tf = (Q/mc) + Ti. This gives me only about 305 K.
     
  2. jcsd
  3. Apr 17, 2014 #2

    rock.freak667

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    In ΔE = Q-W, you aren't accounting for the work done by the system as it is lifting a 10 kg mass a distance of 2 m. How would you find the work done in moving a distance d by a force F?
     
  4. Apr 17, 2014 #3
    Work = a change in energy. I'm not sure how to integrate that into the problem. I've gone over this problem for hours trying to figure out what I'm missing. I could use W = PV, but I don't have enough info. I was trying to use the delta E equation but I'm lost on the steps.
     
  5. Apr 17, 2014 #4
    It should be ΔE=mcΔT, not Q=mcΔT. The latter is correct only if the volume is constant. You need to calculate the work that the gas does in lifting the weight, assuming that the weight is initially at equilibrium and that there is vacuum above the weight. The work the gas does is the change in potential energy of the weight. PΔV=mgΔh.

    Chet
     
  6. Apr 18, 2014 #5
    This must involve a substitution that I'm not catching. The knowns are: heat capacity (c), heat (Q), mass (m), distance (x), initial temp (Ti) and looking for Tf.
    If I have PΔV = mgh, do I need to sub in mcΔT somehow to incorporate c, Q and ΔT? Would anyone be able to show me the work on how this could be written out? Thank you!
     
  7. Apr 18, 2014 #6
    mcΔT=Q-W=Q-mgh
     
  8. Apr 18, 2014 #7

    Andrew Mason

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    In order to do this problem correctly you have to be able to determine how much work is done. Since the problem does not give you the cross-sectional area of the cylinder, you have to assume that the space above the piston inside the cylinder is a vacuum. So the only work done is, as Chestermiller says, mgΔh (using W = PΔV, W = (mg/A)ΔV = mgΔh). [Otherwise P = (mg/A + Patm). So the work done would be PΔV = (mg/A + Patm)ΔV = mgΔh + PatmAΔh].

    Since W = mgΔh and, as you have noted, W = Q-ΔU you should be able to determine ΔU. Then it is just a matter of determining ΔT from ΔU (or ΔE as you have it - the change in internal energy). How is ΔU related to n and ΔT?

    AM
     
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